Homework Set 5 Solutions

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1 MATH Vector Spaces Prof. D. A. Edwards Due: Oct. 7, 2014 Homework Set 5 Solutions 1. Let S, T L(V, V finite-dimensional. (a (5 points Prove that ST and T S have the same eigenvalues. Solution. Let z be an eigenvector for ST corresponding to λ. Then (T ST z = T (ST z = T (λz = λ(t z, so T z is an eigenvector for T S if T z 0. If T z = 0, we see that λ = 0 must be an eigenvalue for ST because ST z = S0 = 0 = 0z. If N (S {0}, then for any y N (S, we have T Sy = T 0 = 0 = 0y, so λ = 0 is an eigenvalue for T S. If S is invertible, let y = S 1 z. Then we have T Sy = T z = 0 = 0y, so λ = 0 is an eigenvalue for T S. Since the roles of T and S can be switched arbitrarily, we have that ST and T S have the same eigenvalues. (b (4 points Show that ST T S I for any S, T L(V. Explain why it might be true if V is infinite-dimensional. Solution. Assume that ST T S = I, and let z be an eigenvector for ST corresponding to λ, the eigenvalue for ST (and hence also T S with smallest real part. Then (ST T Sz = Iz (T Sz = λz z, and hence we see that z is an eigenvector for T S corresponding to λ 1. But this contradicts λ having the smallest real part, so ST T S I. Note that if V is infinite-dimensional, it may have an infinite set of eigenvalues, so there may not be a smallest one. 2. (3 points Let T L(V be an invertible transformation, and let U be a subspace of V. Prove that if U is invariant under T, U is invariant under T 1. Solution. Let B U = {u i } n 1 be a basis for U, and let v U be defined by v = c i u i. If U is invariant under T, then T (v = T c i u i = c i T (u i = d i u i for some constants d i. Since T is invertible, it is onto, so every u U can be written as Copyright 2014 D. A. Edwards All Rights Reserved

2 M672F14Sol5.2 But then u = T 1 (u = T 1 d i u i = and hence U is invariant under T 1 as well. d i u i. T 1 (c i T (u i = c i u i U, 3. An operator N L(V is called nilpotent if N k = O for some integer k. (a (3 points Show that the only eigenvalue for N is λ = 0. Solution. Let z be an eigenvector for N. Then Nz = λz N 2 z = N(Nz = N(λz = λnz = λ 2 z N k z = 0z = 0 = λ k z. Since z 0, λ k = 0 and hence λ = 0. (b (3 points Let P n. Give an example of a nilpotent operator N such that N n+1 = O, but N j O for any j n. Solution. Let N be differentiation. Then applying N repeatedly, we have n 1 N(p(x = N a i x i = (i + 1a i x i n j N j (p(x = (i + j(i + j 1 (1a i x i N n (p(x = n(n 1 (1a 0 x 0 N n+1 (p(x = 0 Therefore, we see that N j 0 for any j n, but also N n+1 = Let T L(V, where U i is invariant under T. Show that (a (3 points Solution. Since R(T = R (T

3 M672F14Sol5.3 each element v V can be written as v = u i, u i (A for a unique choice of vectors u i, so every T v R(T can be written as T v = T u i, Since the U i are invariant under T, T u i and hence T u i R(T. Moreover, since we see that T u i T u j for any i j, so the representation is unique. Since every vector in R(T can be written uniquely as a sum of vectors in R(T, R(T = R(T. (b (3 points N (T = N (T Solution. Let v N (T. Applying T to both sides of (A, we obtain T v = 0 = T u i = w i, w i since the U i are invariant under T. But then by (A with v = 0, we see that w i = 0 by the definition of the direct sum. Thus T u i = 0, which implies that u i N (T. Since every v N (T can be written uniquely as a sum of these u i, we have that N (T = N (T. 5. (8 points Two operators S and T L(V are said to be simultaneously diagonalizable if M(T, Z, Z and M(S, Z, Z are diagonal for the same basis Z of V. Prove that if S and T are simultaneously diagonalizable, ST = T S.

4 M672F14Sol5.4 Solution. From notes in class, we know that Z = {z j } n 1 is an eigenvector basis for V. Denote the entries of M(T, Z, Z by t ij and the entries of M(S, Z, Z by s ij. Defining v V by v = c j z j, we obtain (ST v = S c j T (z j = S = c j t jj n s ij z i = c j n c j t jj s jj z j, t ij z i = S c j t jj z j = c j t jj S(z j (A.1 where we have used the fact that S and T are diagonal, so only s ii and t ii are nonzero. Similarly, n (T Sv = c j S(z j = T c j s ij z i = T c j s jj z j = c j s jj T (z j = c j s jj n t ij z i = c j s jj t jj z j. Since equations (A agree and v is arbitrary, ST = T S. (A.2 6. (8 points Let T L(V, dim n, λ be an eigenvalue of T. Furthermore, assume that the dimension of the eigenspace of T corresponding to λ is m. Show that there exists a basis W for V such that ( λi B M(T, W, W =, B R m (n m, C R (n m (n m, O C where I is the identity matrix in R m m, and O is the zero matrix in R (n m m. Solution. Let B W = {w j } m 1 be a basis for the eigenspace of T corresponding to λ. Extend it to a basis W for V by adding S = {s j } n m+1. Then since T (w j = λw j, we have T (w j = m m ij w i + i=m+1 m ij s j = λw j + m 0w i + i j i=m+1 0s i, where m ij is the ijth entry of M(T, W, W. Since m ij = 0 for i j in the first m columns of M(T, W, W and m ii = λ, these columns form ( λi, O

5 M672F14Sol5.5 where I is the identity matrix in R m m, and O is the zero matrix in R n m m. The other entries are arbitrary, so as desired. ( λi B M(T, W, W = O C, B R m (n m, C R (n m (n m,